Given that integral from 0 to (1/2) dx/(sqrt(1-x²))=pi/6
Given that $\displaystyle\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{6}$, apply Simpson's formula with $n=4$ to approximate this integral and hence $\pi$, and estimate the error.
Explanation of Method
Since $\int dx/\sqrt{1-x^2}=\arcsin x$, the stated identity follows immediately from $\arcsin(1/2)=\pi/6$. Approximating the integral with Simpson's rule and multiplying by $6$ then gives an approximation of $\pi$, whose accuracy is governed by the usual Simpson error bound.
Step 1: Confirm the identity
\[\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}=\big[\arcsin x\big]_0^{1/2}=\arcsin\frac12-0=\frac{\pi}{6}\quad\Longrightarrow\quad \pi=6\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}.\]
Step 2: Simpson's formula, $n=4$ ($2n=8$ intervals), $h=1/16$
Nodes $x_i=i/16$ for $i=0,\dots,8$, function $f(x)=1/\sqrt{1-x^2}$:
| $i$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| $x_i$ | 0 | 1/16 | 1/8 | 3/16 | 1/4 | 5/16 | 3/8 | 7/16 | 1/2 |
| $f(x_i)$ | 1.000000 | 1.001959 | 1.007905 | 1.018056 | 1.032796 | 1.052723 | 1.078720 | 1.112077 | 1.154701 |
| weight | 1 | 4 | 2 | 4 | 2 | 4 | 2 | 4 | 1 |
Weighted sum $S\approx25.132798$, so
\[I_{\text{Simpson}}=\frac{1/16}{3}\times25.132798\approx0.523600.\]
Step 3: Recover $\pi$
\[\pi_{\text{approx}}=6\times0.523600\approx3.141600.\]
\[\boxed{\pi\approx3.1416}\]
Step 4: Accuracy check
\[E=|3.141593-3.141600|\approx7.1\times10^{-6},\]
which sits comfortably alongside the theoretical Simpson error bound for this problem (of similar order, a few$\times10^{-6}$ once carried through to $\pi$), consistent with $f(x)=1/\sqrt{1-x^2}$ being smooth (though increasingly steep) on $[0,1/2]$, well away from its singularity at $x=1$.
Result
Final answer:
\[\boxed{\pi\approx3.1416\ \left(\text{via }\pi=6\int_0^{1/2}\frac{dx}{\sqrt{1-x^2}}\right),\qquad E\approx7\times10^{-6}}\]